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| 1 | +/* |
| 2 | + * Title: 3002. Result Array After Splitting |
| 3 | + * Link: https://leetcode.com/problems/result-array-after-splitting/ |
| 4 | + * |
| 5 | + * Problem: |
| 6 | + * You are given an integer array nums. |
| 7 | + * You must split it into two arrays list1 and list2 such that: |
| 8 | + * - list1 starts with nums[0] |
| 9 | + * - list2 starts with nums[1] |
| 10 | + * - For every i from 2 to n-1, compare the last elements of list1 and list2: |
| 11 | + * -> If last element of list1 > last element of list2, put nums[i] in list1 |
| 12 | + * -> Otherwise, put nums[i] in list2 |
| 13 | + * Finally, return the concatenation of list1 followed by list2. |
| 14 | + * |
| 15 | + * Approach: |
| 16 | + * - Initialize list1 with nums[0] and list2 with nums[1]. |
| 17 | + * - Traverse nums from index 2. |
| 18 | + * - At each step, compare the last element of list1 and list2. |
| 19 | + * - Place nums[i] into the list whose last element is greater (list1) or smaller (list2). |
| 20 | + * - After traversal, build the final result array by appending list1 followed by list2. |
| 21 | + * |
| 22 | + * Dry Run: |
| 23 | + * Example: nums = [2, 1, 3, 3] |
| 24 | + * list1 = [2], list2 = [1] |
| 25 | + * i=2 → compare lastList1=2, lastList2=1 → list1 wins → list1=[2,3] |
| 26 | + * i=3 → compare lastList1=3, lastList2=1 → list1 wins → list1=[2,3,3] |
| 27 | + * Final result = [2,3,3,1] |
| 28 | + * |
| 29 | + * Time Complexity: O(n) → single pass over nums, building result in linear time |
| 30 | + * Space Complexity: O(n) → extra space used for list1 and list2 |
| 31 | + */ |
| 32 | + |
| 33 | +class Solution { |
| 34 | + public int[] resultArray(int[] nums) { |
| 35 | + int length = nums.length; |
| 36 | + |
| 37 | + List<Integer> list1 = new ArrayList<>(); |
| 38 | + List<Integer> list2 = new ArrayList<>(); |
| 39 | + |
| 40 | + list1.add(nums[0]); |
| 41 | + list2.add(nums[1]); |
| 42 | + |
| 43 | + for (int i = 2; i < length; i++) { |
| 44 | + int lastList1 = list1.get(list1.size() - 1); |
| 45 | + int lastList2 = list2.get(list2.size() - 1); |
| 46 | + |
| 47 | + if (lastList1 > lastList2) { |
| 48 | + list1.add(nums[i]); |
| 49 | + } else { |
| 50 | + list2.add(nums[i]); |
| 51 | + } |
| 52 | + } |
| 53 | + |
| 54 | + int[] ans = new int[length]; |
| 55 | + int index = 0; |
| 56 | + for (int ele : list1) { |
| 57 | + ans[index++] = ele; |
| 58 | + } |
| 59 | + for (int ele : list2) { |
| 60 | + ans[index++] = ele; |
| 61 | + } |
| 62 | + return ans; |
| 63 | + } |
| 64 | +} |
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